Find a peak element in this matrix. Return the index of the peak. Guarantee that there is at least one peak, andifthere are multiple peaks,returnany one of them. Example Example 1: Input: [ [1, 2, 3, 6, 5], [16,41,23,22, 6], [15,17,24,21, 7], [14,18,19,20,10], ...
当nums[mid-1]>nums[mid]时,例如3,2,1 我们把R置为mid-1, 反之,则把L置为mid classSolution {publicintfindPeakElement(int[] nums) {if(nums.length==0 || nums==null)return0;intL = 0;intR = nums.length - 1;while(L<R){intmid = (L+R+1)>>>1;if(nums[mid-1]>nums[mid]){ R=...
A peak element is an element that is greater than its neighbors. Given an input array wherenum[i]≠ num[i+1], find a peak element and return its index. The array may contain multiple peaks, in that case return the index to any one of the peaks is fine. You may imagine thatnum[-...
1 public class Solution { 2 public int findPeakElement(int[] num) { 3 if(num.length == 0 || num.length == 1) 4 return 0; 5 boolean isIncr = false; //是否为升序 6 int index = -1; //peak索引值 7 int numArray[] = new int[num.length + 2]; 8 for(int i = 0; i < ...
public int findPeakElement(int[] nums) { if (nums == null || nums.length == 0) return -1; for (int i = nums.length - 2; i >= 0; i--) { if (nums[i] > nums[i + 1]) continue; else { return i + 1; } }
class FindPeakElement { fun findPeakElement(nums: IntArray): Int { return findPeakElement(nums, 0, nums.lastIndex) } private fun findPeakElement(nums: IntArray, startIndex: Int, endIndex: Int): Int { val midIndex = (endIndex + startIndex) / 2 val midVal = nums[midIndex] val left...
Can you solve this real interview question? Find a Peak Element II - A peak element in a 2D grid is an element that is strictly greater than all of its adjacent neighbors to the left, right, top, and bottom. Given a 0-indexed m x n matrix mat where no t
Find Peak Element I A peak element is an element that is greater than its neighbors. Given an input array where num[i] ≠ num[i+1], find a peak element and return its index. The array may contain multiple peaks, in that case return the index to any one of the peaks is fine. ...
peaks = [a[x for pprevX,ax,nnextX in zip(L[:1]+L,L,L[1:]+L[-1:]) if p<=a>=n]prevX <= x >= nextX] # [3, 5, 2] Use zip to form tuples of each item in L with the previous and next items (padding with the first and last as needed) peaks = [a for p,a...
这样就可以找到一个峰值了. 代码 class Solution {public: int findPeakElement(vector& nums) { int left=0; int right=nums.size()-1; while(left 参考文献 [LeetCode] Find Peak Element 求数组的局部峰值]